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Friday puzzle -- Global circumnavigation
A fleet of identical planes is based on a small island. Each plane has the fuel capacity to fly half-way around the world. These planes have the ability to transfer any amount of fuel in flight. The base island is the only source of fuel. All planes fly at the same speed and direction reversals are instantaneous.
How many planes will be required to allow one of them to circumnavigate the globe, with all of them returning safely to base?
Re: Friday puzzle -- Global circumnavigation
Assuming that fueling operations (both in-flight and on the ground) are also instantaneous, you can do it with a 'magnificent' number of planes.
Re: Friday puzzle -- Global circumnavigation
My planes can do it with five.
Lee
Yes, but
not optimal if your reference is to Yul Brynner and Steve McQueen.
I think it's 5 too?
I thought I could do it with 4, but kept crashing in the ocean
I think 4 can work
Nope, reread and I think I can do it in three
with two planes in the ocean... (since the conditions don't seem to require that all planes return)
two planes fly 1/4 of the way, when plane #1 refuels with Plane #2's remaining fuel. this enables it to go 3/4 of the way. Plane three heads out in the opposite direction and meets #1 at the 3/4 mark, having used 1/4 of it's fuel to get there.
Transferring it's remaining fuel completes the last 1/4.
Only two pilots die, but hey, the one you asked about makes it.
Re: Nope, reread and I think I can do it in three
It does require that all return safely. ;-)
Best I can do is 6
Two gets you 2/3 of the way, but 4 more are needed to bring them all home.
On second thought, make that 4 planes with 6 loads of fuel.
since we know WC doesn't have an edit feature :-)
I must have not read it to the end...
or
Some needs to tell Alex that they hacked his original post to include that in the requirements.
It was so much easier when we didn't care about the pilots ...
Sigh.
S
Re: since we know WC doesn't have an edit feature
;-)
I know, he always puts on some kind of a constraint that makes it hard.;-)
The goal
Sorry I was out of pocket yesterday, but if anyone is still following,
Steve got the right answer, but you need to save the pilots.
Ed, Lee, and Bll, why don't you post your solutions for 6, 5, and 4. Collectively, you might have the ideas that when combined may get you to the three plane solution.
Re: The goal
Not much time to work on this at the moment. I spent all Saturday coaching my son's robotics team to a first-place finish at the regional qualifiers. Now we need to prepare for the State Championship next weekend.
Breifly, the 4 plane solution is:
* 4 planes start in the same direction with full tanks.
* At 1/8th of the way (with 1/4 tank consumed), plane A transfers 1/4 tank each to planes B & C and returns to base.
* At 1/4 of the way, planes B & C transfer 1/4 tank each to plane D and return to base.
* Plane D (now with a full tank) continues on.
* When planes B & C arrive for an instant fill-up, they instantly depart with plane A in the opposite direction.
* Plane A transfers 1/4 tank each to B & C at 1/8th as before.
* Planes B & C meet plane D at the 3/4 mark and transfer 1/4 tank each so they all have 1/2 tank to get home.
In this scheme, plane A is only 50% utilized, so a more optimal solution probably exists.
On the right track
And congrats on your son's robotics team!
Hint
Follow Bill's approach as far as you can, but with no plane D.
The beginning and end of the flight are symmetric, with A and B switching roles.
Re: Hint
I see it now. The fourth plane is not necessary. Plane A can top off B and C at the 1/8 mark. Then Plane B can top off C at the 1/4 mark. Then just reverse the process on the home stretch.
:-)
Re: Hint
That's what I was doing, but I assumed that any plane leaving the start point was a new plane.....
Lee